Chapter 3

Transportation Problems

QEM - Chapter 3

Design of Transport Networks

Situation: Location of warehouses and customers are given Supply and demand given

Example: 3 warehouses and 4 customers Transportation cost per unit from i to j Total demand must be = total supply

Factory or customer

warehouse V1 V2 V3 V4 Production

F1 10 5 6 11 25

F2 2 2 7 4 25

F3 9 1 4 8 50

Demand 15 20 30 35 100

Transportation Problem: Model & LP

(c) Prof. Richard F. Hartl Kapitel 3 / 2

(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/3

Solution in the Uncapacitated case

No capacity constraint solution using column minimum procedure

customer

Factory V1 V2 V3 V4 Capacity used

F1 10 5 6 11

F2 1 2 7 4

F3 9 1 4 8

Demand 15 20 30 35

15

20 30

35 15 + 35 = 50

20 + 30 = 50

Total cost = 15 * 1 + 20 * 1 + 30 * 4 + 35 * 4 = 295

(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/4

Capacity constraints

Not all customers will be delivered from their closest factory!

Solution as a special LP problem- starting (basic feasible) solution- iteration (modi, stepping stone)

(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/5

Capacity constraints

Total capacity = total demand

Customer

Factory V1 V2 V3 V4 capacity

F1 10 5 6 11 25

F2 1 2 7 4 25

F3 9 1 4 8 50

Demand 15 20 30 35 100

100

100

(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/6

Constructive Heuristics (Basic Feasible Solution)

Heuristics: Simple to apply, Simple to understand „Reasonably“ good solutions Optimality not guaranteed

Examples:Column minimum („Greedy“)

Vogel approximation („Regret“)

(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/7

Column Minimum Algorithm

0. Initialization: empty transport tableau. No rows or columns are deleted

1. Proceed with the first (not yet deleted) column (from left to right)

2. In choose the cell with the smallest unit cost cij and choose

transportation quantity xij maximal.

3. If column resource depleted delete column j,

ORif row resource depleted delete row i.

4. Just one row or columd not deleted all cells in this row or colum get maximal transportation quantityOtherwise continue with 1.

EXAMPLE

(c) Prof. Richard F. Hartl QEM - Chapter 3 Kapitel 3/8

Column Minimum Algorithm

i / j 1 2 3 4 Capacity

1 10 5 6 11 25

2 1 2 7 4 25

3 9 1 4 8 50

Demand 15 20 30 35 100

15

20 30 0

10

25

30

10

0

0

0

Total cost = 15 * 1 + 20 * 1 + 30 * 4 + 25 * 11 + 10 * 4 + 0 * 8 = 470

Just one colums not deleted

ALGORITHM

SS 2011 EK Produktion & LogistikKapitel 3/9

Vogel Approximation

0. Initialization: empty transport tableau. No rows or columns are deleted

1. In each row and column (not yet deleted) compute opportunity cost = difference between the smallest and the second smallest cij (not yet deleted).

2. Where this opportunity cost is largest choose smallest cij and make xij maximal.

3. If column resource depleted delete column j,

ORif row resource depleted delete row i.

4. Just one row or columd not deleted all cells in this row or colum get maximal transportation quantityOtherwise continue with 1.

SS 2011 EK Produktion & Logistik Kapitel 3/10

Vogel Approximation

i / j 1 2 3 4 capacity

1 10 5 6 11 25

2 1 2 7 4 25

3 9 1 4 8 50

Nachfrage 15 20 30 35 100

15

20 5 25

10

25

8 21

3

1

1

4

10 2

25

4 3

30 4

5

5

Total cost = 15 * 1 + 20 * 1 + 25 * 6 + 5 * 4 + 10 * 4 + 25 * 8

= 445

Column deleted recompute row differences

Row deleted recompute column differences

Jut one row not deleted

(besser als 470 zuvor)

QEM - Chapter 3

General formulation:

m supplyers with supply si, i = 1, …, m

n customers with demand dj, j = 1, …, n

Transportation cost cij per unit from i to j, i = 1, …, m; j = 1, …, n

variable: Transportation quantity xij from i to j

LP-Formulation:

Transportation cost

Supply

Demand

Non negativity

min1 1

m

iij

n

jijxcK

n

jiji xs

1i = 1, …, m

m

iijj xd

1j = 1, …, n

0ijx i = 1, …, m; j = 1, …, n

Mdn

jj

1

m

1=iisData must satisfy Total demand = Total supply

(c) Prof. Richard F. Hartl Kapitel 3 / 11

QEM - Chapter 3

In example:

Supply: x11 + x12 + x13 + x14 = 25 (i=1)

x21 + x22 + x23 + x24 = 25 (i=2)

x31 + x3 2+ x33 + x34 = 50 (i=3) :

Damand: x11 + x21 + x31 = 15

(j=1) x12 + x22 + x32 = 20

(j=2) x13 + x23 + x33 = 30 (j=3)

x14 + x24 + x34 = 35 (j=4) Non negativity:

xij 0 für i = 1, … , 3; j = 1, … , 4

K = (10x11+5x12+6x13+11x14) + (x21+2x22+7x23+4x24) + (9x31+x32+4x33+8x34) min

(c) Prof. Richard F. Hartl Kapitel 3 / 12

QEM - Chapter 3

Solution: As LP (o.k. but less efficient)

In each column exactly 2 of the m + n elements are ≠ 0

Make use of special structure:

1 . . . 1

1 . . . 1

. . .

1 . . . 1

1 ..

.

1

1 ..

.

1

. . . 1 ..

.

1

Transportation simplex or network simplex

Starting solution

Iteration (stepping stone)

(c) Prof. Richard F. Hartl Kapitel 3 / 13

QEM - Chapter 3

Exakt Algorithm: MODI, Stepping Stone Equivalent to simplex method

Start with basic feasible solution (m+n-1 basic variables) Iteration step:

Initalize the tableau

i\j 1 2 … n si ui

1c11 c12

…c1n

s1 u1

2c21 c22

…c2n

s2 u2

… … … … … …

mcm1 cm2

…cmn

sm um

dj d1 d2 … dn

vj v1 v2 … vn

(c) Prof. Richard F. Hartl Kapitel 3 / 14

QEM - Chapter 3

For all non basic variables compute cij – ui – vj (reduced cost coefficient in objective function). New (entering) BV where this coefficient is most negative. If all coefficients non-negative optimal solution.

Compute dual varables ui und vj [MODI]

cij = ui + vj if xij is basic variable

Since ui and vj are not unique, normalize one of these dual varables = 0. (Choose row or colums with most basic varaibles easier computation of ui and vj)

Increase new BV and perform chain reaction (donor cell, recipient cell). Transportation quantities in each row and column must remain same. The BV, which first becomes zero, leaves the basis. [stepping stone]

Compute new basic solution (perform chain reaction).

(c) Prof. Richard F. Hartl Kapitel 3 / 15

QEM - Chapter 3

Example

i\j 1 2 3 4 si ui

110 5 6 11

25

21 2 7 4

25

39 1 4 8

50

dj 15 20 30 35

vj

15 10

10 15

15 35

-3-4

-7-6

25

10 5 10 14

-3

-6

0

For didactical raesons we choose poor starting solution, because after Vogel only few steps (if any) can be demonstrated

(c) Prof. Richard F. Hartl 16

QEM - Chapter 3

Total cost = 10*15 + 5*10 + 2*10 + 7*15 + 4*15 + 8*35 = 665.

Check equality of primal and dual objective (optional):

The most negative cost ceofficient cij – ui – vj of a NBV is -7 for x24 new basic variable x24.

Chain rection: increase new BV (starting from 0) by investigate consequences for other BV (in some row and some column must be subtracted) The BV, that constrains most is the leaving BV.

j

n

jj

m

iii

m

iij

n

jij dvsuxcK

111 1

K = 25*0 + 25*(-3) + 50*(-6) + 15*10 + 20*5 + 30*10 + 35*14 = 665

(c) Prof. Richard F. Hartl Kapitel 3 / 17

QEM - Chapter 3

Chain reaction:

i\j 1 2 3 4 si ui

110 5 6 11

25

21 2 7 4

25

39 1 4 8

50

dj 15 20 30 35

vj

15+

15-

35-

15 10

10

-3-4

-7-6

25

10 5 10 14

-3

-6

0

New BV x24 increases from 0 to . Add + odor - for some other BV.

If x24 increases by , x23 and x34 will decrease by , and x33 increases by .For = 15 BV x23 becomes 0 → BV x23 leaves.

+

K = 665 – 7 *

= 665 – 7*15 = 560

(c) Prof. Richard F. Hartl Kapitel 3 / 18

QEM - Chapter 3

New BV x24 gets value = 15 → chain reaction x34 = 35-15 = 20

x33 = 15+15 = 30

x23 is no BV anymore, all other BV remain same

i\j 1 2 3 4 si ui

110 5 6 11

25

21 2 7 4

25

39 1 4 8

50

dj 15 20 30 35

vj

43

7-6

-5-2

10+ 15-

10-

10 5 3 7

-3

1

0

30 20

15

Next iteration:

K = 560 – 6 *

= 560 – 6*10 = 500

(c) Prof. Richard F. Hartl Kapitel 3 / 19

QEM - Chapter 3

Next iteration:

i\j 1 2 3 4 si ui

110 5 6 11

25

21 2 7 4

25

39 1 4 8

50

dj 15 20 30 35

vj

-2-3

76

14

20

10 5 9 13

-9

-5

05-

30-

15-

20+

10+

K = 500 – 3 *

= 500 – 3*5 = 485

(c) Prof. Richard F. Hartl Kapitel 3 / 20

QEM - Chapter 3

Next iteration:

i\j 1 2 3 4 si ui

110 5 6 11

25

21 2 7 4

25

39 1 4 8

50

dj 15 20 30 35

vj

-2

13

73

4

20- 5+

7 5 6 10

-6

-2

0

25-

10

25

15

K = 485 – 2 *

= 485 – 2*20 = 445

(c) Prof. Richard F. Hartl Kapitel 3 / 21

QEM - Chapter 3

Next iteration:

i\j 1 2 3 4 si ui

110 5 6 11

25

21 2 7 4

25

39 1 4 8

50

dj 15 20 30 35

vj

2 13

75

4

25

5 1 4 8

-4

0

2

20 5

10

25

15

All coefficients non-negative → optimal solution found.

Basic variables: x13 = 25

x21 = 15

x24 = 10

x32 = 20

x33 = 5

x34 = 25

Total cost: K = 445

(c) Prof. Richard F. Hartl Kapitel 3 / 22

QEM - Chapter 3

TP is a LP-Problem with equality constraints dual variables ui azw. vj can also be negative (free variables).

Sensitivity analysis

If small enough, then the dual variables ui and vj do not change cost change by (ui + vj):

From duality theory follows

Clearly some si AND soem dj must change at the same time;Otherwise total demand ≠ total capacity

si si + for some i anddj dj + for some j

K K + (ui + vj)

(c) Prof. Richard F. Hartl Kapitel 3 / 23

QEM - Chapter 3

Example:

i\j 1 2 3 4 si ui

11 4 2 5

22

26 1 5 6

24

37 5 3 5

16

dj 10 13 22 17

vj

10

1 -1 2 4

2

1

0

10 6

11

12

13

Change of data?

s1 s1 + and

d2 d2 +

Optimal cost change to K = 173 + (u1 + v2) = 173 - ; i.e. cost decrease! (this paradoxon can occur if some ui and/or vj are negative. In most cases cost will increase.)

22 +

13 +

(c) Prof. Richard F. Hartl Kapitel 3 / 24

QEM - Chapter 3

How large can be before basis changes: similar to stepping stone

i\j 1 2 3 4 si ui

11 4 2 5 22

+

26 1 5 6

24

37 5 3 5

16

dj 10 13 + 22 17

vj

10

1 -1 2 4

2

1

0

6+10-

11-

12+

13+

K = 173 -

Clearly x33 first becomes 0, if increases. Hence upper bound 10. Similar for negative . Here x34 first becomes 0, if decreases. Hence lower bound -6.

Check: K = 1*10+2*(12+ ) + 1*(13+ ) +6*(11- ) +3*(10- ) +5*(6 + ) = 173 -

(c) Prof. Richard F. Hartl Kapitel 3 / 25